\documentclass[10pt]{article} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \usepackage{amsmath,amscd,amsthm,amsfonts,amssymb} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{exer}{Exercise} \theoremstyle{plain} \newtheorem{thm}{Theorem} \newtheorem{lem}{Lemma} \newtheorem{cor}{Corollary} \newtheorem{prop}{Proposition} \renewcommand{\P}{\mathbb{P}^1} \renewcommand{\k}{\kappa} \renewcommand{\H}{\mathbb{H}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\G}{\Gamma} \renewcommand{\O}{\mathcal{O}} \newcommand{\GO}[1]{\Gamma_0(#1)} \newcommand{\res}[2]{\mathrm{Res}_{#2}(#1)} \newcommand{\Gb}[1]{\bar{\Gamma}(#1)} \newcommand{\SL}[1]{\mathbf{SL}_2(#1)} \newcommand{\PSL}[1]{\mathbf{PSL}_2(#1)} \newcommand{\GL}[1]{\mathbf{GL}_2(#1)} \newcommand{\V}[2]{\bigl[\begin{smallmatrix}#1 \\#2\end{smallmatrix}\bigr]} \newcommand{\z}{\zeta} \newcommand{\zt}[1]{\zeta_{p^{#1}}} \newcommand{\zr}{\zt{r}} \newcommand{\s}{\sigma} \renewcommand{\a}{\alpha} \renewcommand{\b}{\beta} \renewcommand{\c}{\gamma} \newcommand{\m}{\mathfrak{m}} \newcommand{\p}{\mathfrak{p}} \newcommand{\q}{\mathfrak{q}} %\newcommand{\cl}{\mathcal{Cl}} \newcommand{\ga}{\mathfrak{a}} \newcommand{\gc}{\mathfrak{c}} \newcommand{\nsimeq}{\not\simeq} \DeclareMathOperator{\deck}{Deck} \DeclareMathOperator{\aut}{Aut} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\gal}{Gal} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\disc}{disc} \DeclareMathOperator{\tr}{Tr} \DeclareMathOperator{\n}{N} \DeclareMathOperator{\mat}{N} \DeclareMathOperator{\cl}{Cl} \title{Algebraic Number Theory, PS 1} \author{Bryden Cais} \begin{document} \maketitle \begin{enumerate} \item Let $L/K$ be an extension of number fields. It is not true that for $\alpha\in L$, $\tr(\alpha),\n(\alpha)\in \O_K$ implies $\alpha\in \O_L.$ As a counterexample, set $K=\Q$ and let $\alpha$ be a root of $$f(x)=x^3+\frac{1}{3}x-1$$ and let $L=K(\a)$ Then we have $\tr(\a)=0,\n(\a)=1.$ It is clear that $f$ is irreducible: if it were not, it would have a rational root. But such a root would have numerator and denominator dividing 3, which (after a brief search) is impossible. It follows that $f$ is the minimal monic polynomial of $\a$ so that $\a$ is not integral. If $\deg(L/K)=2$ then $\tr_{L/K}(\a),\n_{L/K}(\a)\in\O_K$ if and only if $\a\in\O_L$. One direction is clear. For the reverse, observe that $\a$ is a root of the monic polynomial $$x^2-\tr_{L/K}(\a)x+\n_{L/K}(\a),$$ and is therefore integral over $\O_K$ (and hence over $\Z$). Since $\a\in L,$ we have $\a\in\O_L$. %On the other hand, if $\a\in\O_L,$ then %$\a$ satisfies monic polynomial with coefficients in $\Z$. %If we had $\a\in \O_L,$ then we would have %$$\disc(1,\a,\a^2)\in\Z,$$ %since this latter is integral (being a determinant of integral things) and in $\Q$ (since it is fixed by any embedding %of $L$ into $\C$). Using 9) we find that %\begin{align*} %\disc(1,\a,\a^2)&=\pm \n_{L/K}\left(3\a^2+\frac{1}{3}\right).\\ %\end{align*} %Let the other roots of $f(x)$ be $\b,\c$. Then we have %\begin{align*} % \a+\b+\c&=0\\ % \a\b+\a\c+\b\c&=\frac{1}{3}\\ % \a\b\c&=1, %\end{align*} %and %\begin{align*} % \n_{L/K}\left(3\a^2+\frac{1}{3}\right)&=\n_{L/K}\left(\frac{2\a^3+1}{\a}\right)\\ % &=\left(\frac{(2\a^3+1)(2\b^3+1)(2\c^3+1)}{\a\b\c}\right)\\ % &=8(\a\b\c)^3+2(\a^3+\b^3+\c^3)+4(\a^3\b^3+\a^3\c^3+\b^3\c^3)+1\\ % &=9+2((\a+\b+\c)^3-3(\a^2\b+\b^2\a+\a^2\c+\c^2\a+\b^2\c+\c^2\b)-6\a\b\c)+4((ac+ab+bc)^3-3()) %\end{align*} \item Let $d$ be square free and $p$ a prime not dividing $2d$. Set $K=\Q(\sqrt{d}).$ Since $\Delta_K=d$ or $4d$, we see that $p$ does not ramify in $K$. Now $d$ is not a square modulo $p$ if and only if $x^2-d$ has no solutions in $\Z/(p)$. But in any case, $x^2-d$ has a root in $\O_K$ (obviously) and hence has a root in $\O_K/\wp$ for any prime $\wp$ over $p$. Therefore, $d$ is not a square modulo $p$ if and only if $$\O_K/\wp \nsimeq \Z/(p).$$ Since $\deg(K/\Q)=2,$ it follows that $\O_K/\wp$ is a degree two extension of $\Z/(p)$ and hence that $d$ is not a square modulo $p$ if and only if $p$ remains prime in $K$. \item Let $D$ be a dedekind domain with a finite number of prime ideals. We show that $D$ is a PID. Let $\p_1,\p_1,\ldots,\p_n$ be the distinct prime ideals of $D$. Pick $\a_i\in p_i\setminus p_i^2$ such that $\a\not\in \p_j$ for $j\neq i$. We know that such a choice is always possible by, for example, the chinese remainder theorem (problem 4). Let us look at the prime decomposition of $(\a_i).$ Since $a_i\not\in\p_j$ for $i\neq j,$ we see that $\p_j\not|(\a_i).$ Similarly, since $\a_i\not\in\p_i^2,$ we have $\p_i^2\not| (\a_i)$. It follows that $(\a_i)=p_i,$ that is, each prime ideal is principal. Hence, since $D$ is a dedekind domain and we therefore have unique factorization into prime ideals, every ideal of $D$ is principal. \item Define $$f:\O\longrightarrow \bigoplus_{i=1}^n \O/\ga_i$$ by $$f(x)=(x\mod \ga_1,x\mod \ga_2,\ldots,x\mod \ga_n).$$ We claim that $f$ is surjective. It is sufficient to show that we can obtain the element $(0,\ldots,0,1,0,\ldots,0)$ where the $1$ occurs in the $i^{\text{th}}$ place, for any $i$. Since $\ga_i+\ga_j=\O$ for $i\neq j$, we have $$\ga_i+\bigcap_{j\neq i}\ga_j=\O.$$ To see this, we proceed by induction: let $\a\in \ga_i,\b\in\ga_j$ satisfy $\a+\b=1$. Let $I\subset \O$ be any ideal such that $I+\ga_i=\O$ and $I+\ga_j=\O$, and let $\c\in I,\a^{'}\in \ga_i$ satisfy $\c+\a^{'}=1.$ Then we have $$\a\c+\b\c=\c$$ and hence that $$\a^{'}+\a\c+\b\c=1.$$ It follows that $$\a+\a^{'}\b+\a\c\b+\b^2\c=1.$$ Since $\a+\a^{'}\b\in \ga_i$ and $\a\c\b+\b^2\c\in I\cap\ga_j$, we are done. Therefore, there exists $\c\in \bigcap_{j\neq i}\ga_j$ and $\a\in \ga_i$ such that $\a+\c=1$. Then $\c\in\O$ satisfies $\c\mod \ga_j=0$ for $j\neq i$ and $\c\mod\ga_i=1$. It follows that $f$ is surjective. The kernel of $f$ is the ideal of all $x\in\O$ with $x\in \ga_i$ for all $i$. That is, $$\ker f=\bigcap_{i=1}^n \ga_i:=\ga.$$ We then have the isomorphism $$\O/\ga\simeq\bigoplus_{i=1}^n \O/\ga_i.$$ \item Let $K=\Q(\sqrt{-5})$ and $L=\Q(\sqrt{-6})$. \begin{enumerate} \item We first show that $\O_K$ is not a UFD. Consider the two factorizations $$2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})=6.$$ We claim that $2,3,(1\pm\sqrt{-5})$ are all irreducible elements of $\O_K$. Suppose not. Then, for example, we could write $2=(a+b\sqrt{-5})(c+d\sqrt{-5})$ with $a,b,c,d\in \Z$ since $\O_K=\Z[\sqrt{-5}].$ Taking norms gives $$4=(a^2+5b^2)(c^2+5d^2),$$ from which it follows (since we are assuming our factorization of $2$ to be non trivial) that $a^2+5b^2=c^2+5d^2=2$. But $2$ is not a square modulo $5$ and we have reached a contradiction. Similar considerations show that if $3$ is not irreducible, then $3=a^2+5b^2$ for some $a,b\in\Z$, again a contradiction. Now if $\O_K$ were a UFD then we would have $2|(1\pm\sqrt{-5})$. In either case, this requires that $4|6,$ which is a contradiction. \item We use the Minkowski bound to determine the class numbers of $K,L$. We have \begin{align*} \Delta_K&=-20,\\ \Delta_L&=-24, \end{align*} and $$d_K=d_L=2.$$ Observe that $s=1$ in both cases. The Minkowski bound is then $$\frac{2!}{2^2}\left(\frac{4}{\pi}\right)\sqrt{20}<3$$ for $K$ and $$\frac{2!}{2^2}\left(\frac{4}{\pi}\right)\sqrt{24}<4$$ for $L$. Therefore, every ideal class of $K$ contains an $\O_K$ ideal of norm at most $2$. Therefore, we only need to consider those ideals of $\O_K$ of norm exactly $2$. But any ideal of norm $2$ is a prime above $2$, and we know that $2$ totally ramifies in $K$ since it divides the discriminant. Moreover, since $x^2-5\equiv (x+1)^2\mod 2,$ we have (since $\O_K=\Z[\sqrt{5}]$), $$(2)=(2,1+\sqrt{5})^2.$$ Moreover, it is not difficult to see that $(2,1+\sqrt{-5})$ is not principal: if it were, we could write $2=a^2+5b^2$ for $a,b\in\Z$, which is absurd as we have already seen. Thus, $\cl_{\O_K}$ consists of two elements and is hence isomorphic to $\Z/(2).$ For $L$, we proceed as above and find \begin{align*} (2)&=(2,\sqrt{-6})^2\\ (3)&=(3,\sqrt{-6})^2. \end{align*} As before, it is not hard to see that $(2,\sqrt{-6}),(3,\sqrt{-6})$ are not principal. However, we have $$(2,\sqrt{-6})(3,\sqrt{-6})=(6,2\sqrt{-6},6\sqrt{-6},-6)= (\sqrt{-6}),$$ so that if $[\alpha]$ denotes the ideal class of $\alpha\subset\O_L$, we have $$[(2,\sqrt{-6})]=[(2,\sqrt{-6})]^{-1},$$ from which it follows that $\cl_{\O_L}\simeq \Z/(2).$ \end{enumerate} \item Suppose that $K$ is a norm Euclidean imaginary quadratic field. Then given any $\a,\b\in\O_K,$ there exists $q,r\in \O_K$ such that $$\a=\b q+r$$ and $$\n_{K/\Q}(r)<\n_{K/\Q}(\b).$$ Observe that $\O_K$ is a lattice in $\C$ and that the norm of any $\a\in K$ is just the square of its usual euclidean distance from the origin ($\a=u+v\sqrt{-D}=u+iv\sqrt{D}$ with $u,v\in \Q$, and $\n_{K/\Q}(\a)=u^2+dv^2$ while the distance of $(u,v\sqrt{D})$ from the origin is $u^2+Dv^2$). Thus, $K$ is norm euclidean if and only if given $\a,\b$ as above we can find $q\in\O_K,t\in K$ such that $$\frac{\a}{\b}=q+t$$ where $\n_{K/\Q}(t)<1$. Therefore, $K$ is norm euclidean if and only if given any point $x$ of $\C$ (since any point of $\C$ can be approximated as closely as we like by an element $\a/\b$ of $K$ with $\a,\b\in\O_K$ as above) we can find a lattice point $q\in\O_K$ of distance less than $1$ away from $x$. Equivalently, $K$ is norm euclidean if and only if the minimum distance from any interior point of a fundamental domain to a lattice point on the boundary of that fundamental domain is strictly less than $1$. We then have two cases. \begin{enumerate} \item $-D\not\equiv 1\mod 4:$ In this case, $\O_K=\Z\oplus\Z\sqrt{-D}.$ A fundamental cell $F$ for the lattice is the rectangle of side lengths $1,\sqrt{D}$. Given a point $p=x+iy\in F$, let $d(p)=\min_{q\in\O_K}|p-q|^2$, where $|*|$ denotes the usual complex absolute value. Then it is obvious that $$\sup_{p\in F}d(p)$$ occurs when $p=1/2+\sqrt{-D}/2$. In this case, $p$ is equidistant from the four neighboring lattice points and the distance to any of them is $$\frac{\sqrt{1+D}}{2}.$$ This is less than $1$ if and only if $D=1$ or 2. \item $-D\equiv 1\mod 4:$ Here, $\O_K=\Z\oplus\Z\left(\frac{1+\sqrt{-D}}{2}\right).$ A fundamental cell $F$ for the lattice is the parallelogram with vertices $$\left\{(0,0),(1,0),(1/2,\sqrt{D}/2),(3/2,\sqrt{D}/2)\right\}.$$ Let $d(p)$ be as above. Again, it is not difficult to see that $$\sup_{p\in F}d(p)$$ occurs when $p=3/4+\sqrt{D}/4$. In this case, we have $$d(P)=\frac{1+D}{16}.$$ It follows that we must have $D<15$. Since we are requiring that $-D\equiv 1(4)$ the only possible choices are $$D=3,7,11.$$ \end{enumerate} Thus, the only norm euclidean imaginary quadratic fields are $$\Q(\sqrt{-1}),\Q(\sqrt{-2}),\Q(\sqrt{-3}),\Q(\sqrt{-7}),\Q(\sqrt{-11}).$$ It is not true that every quadratic imaginary field with class number one is norm euclidean. For example, $\Q(\sqrt{-19})$ has class number one but is not norm euclidean as the following counterexample shows: $$\a=3+\sqrt{-19},\b=4.$$ In this case, the closest lattice points to $\a/\b$ are $1$ and $\frac{1+\sqrt{-19}}{2}$. But in one case we have $$\n_{K/\Q}(\a-\b)=20\nless \n_{K/\Q}(\b)=16$$ while in the other, $$\n_{K/\Q}\left(\a-\b\left(\frac{1+\sqrt{-19}}{2}\right)\right)=\n_{K/\Q}(1-\sqrt{-D})=20\nless \n_{K/\Q}(\b)=16.$$ \item Let $\O$ be a dedekind domain and $\ga$ a nonzero ideal of $\O$. Suppose that $$\ga=\p_1^{a_1}\ldots \p_r^{a_r}.$$ Then by the chinese remainder theorem we have $$\O/\ga\simeq \bigoplus_{i=1}^r \O/\p_i^{a_i}.$$ Suppose that every ideal of $\O/\p_i^{\a_i}$ is principal and let $I\subset \O/\ga$ be an ideal of $\O/\ga$. Then $I$ corresponds to an ideal $\tilde{I}\subset \O$ containing $\ga$. Clearly, $\tilde{I}/\p_i^{a_i}$ is an ideal of $\O/\p_i^{a_i}$ and is principal by assumption, say $$\O/\p_i^{a_i}=x_i\O/\p_i^{a_i}.$$ Then we see that $I$ is a principal ideal of $\O/\ga$ generated by $$(x_1,x_2,\ldots,x_r)\in \bigoplus_{i=1}^r \O/\p_i^{a_i}\simeq \O/\ga.$$ It therefore suffices to show that for any prime ideal $\p,$ every ideal of $\O/\p^m$ is principal. Let $I$ be any ideal of $\O/\p^m$. Then $I$ corresponds to an ideal $\tilde{I}$ of $\O$ containing $\p^m$. But since $\p^m\subset \tilde{I}$ we have $\tilde{I}|\p^m$ and hence that $\tilde{I}=\p^n$ for some $n\leq m$. If $n=m$ then $I$ is the zero ideal and obviously principal. Otherwise we may assume that $n0$. We distinguish two cases: \begin{enumerate} \item $p=q$. Since $p$ totally ramifies in $E_r$ for all $r$, there is exactly one prime above $p$. In this case, we see that $\p$ splits into at most $\deg(K/Q)$ primes in $L_r$ for any $r$. \item $p\neq q$. Let $\q$ be a prime over $q$ in $E_r$. Recall from lecture that $E_r/\Q$ is Galois and that $$e(\q/q)=1$$ and $$f(\q/q)=f_r$$ is the order of $q$ modulo $p^r$. Since $e_rf_rd_r=\varphi(p^r)$, where $d_r$ is the number of primes lying over $q$ in $E_r$, we see that $$d_r=\frac{p^{r-1}(p-1)}{f_r}$$ for any $r$. Now let $r_0$ be such that $$q^{f_1}=1+cp^{r_0},$$ where $p\not| c$. Then $f_1=f_r$ for $1\leq r\leq r_0$. Observe that $$q^{f_1p^{j-1}}\equiv 1+cp^{r_0+j-1}\mod p^{r_0+j}\not\equiv 1\mod p^{r_0+j}$$ but $$q^{f_1p^{j}}\equiv 1+cp^{r_0+j}\mod p^{r_0+j}\equiv 1\mod p^{r_0+j}$$ for any $j\geq 1$. It follows that $f_{r_0+j}|f_1p^j$ but $f_{r_0+j}\not|f_1p^{j-1}$, and hence (since $f_1|f_r$ for any $r\geq 1$) that $$f_{r_0+j}=f_1p^j.$$ Therefore, $$d_{r_0+j}=\frac{p^{r_0+j-1}(p-1)}{f_{1}p^j}=\frac{p^{r_0-1}(p-1)}{f_{1}}$$ for all $j\geq 1$. This proves our claim. Putting this analysis together, it is not difficult to see that $$c(\p)=\frac{p^{r_0-1}(p-1)}{f_1}\deg(K/Q),$$ where $f_1$ is the order of $l=\p\cap\Z$ modulo $p$ and $r_0$ is such that $$l^{f_1}\equiv 1\mod p^{r_0}$$ but $$l^{f_1}\not\equiv 1\mod p^{r_0+1}.$$ \end{enumerate} \item Let $L=K[\a]$ be a finite, separable extension and let $f$ be the minimal monic polynomial of $\a$ over $K$. Set $n=\deg(f)$ and let $\sigma_i$ for $i=1,\ldots,n$ be the $n$ embeddings of $L$ in $\C$. We have \begin{align*} \Delta(1,\a,\a^2,\ldots,\a^{n-1})&=\begin{vmatrix}1 & \a & \a^2 &\cdots & \a^{n-1}\\ 1 & \sigma_1(\a) & \sigma_1(\a)^2 &\cdots & \sigma_1(\a)^{n-1}\\ \vdots & & &\cdots & \vdots \\1 & \sigma_n(\a) & \sigma_n(\a)^2 &\cdots & \sigma_n(\a)^{n-1}\end{vmatrix}^2\\ &=\prod_{1\leq i